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sloppycanada

  • one year ago

Find f(g(x)) if f(x) = 2x + 1 and g(x) = x2 - 1.

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  1. Jhannybean
    • one year ago
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    Wherever you see an x in your function for f(x) replace it with the function of g(x). \[f(g(x)) = 2(x^2-2)+1\]

  2. anonymous
    • one year ago
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    did u understood?

  3. sloppycanada
    • one year ago
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    \[2x^2-3\] @Jhannybean

  4. sloppycanada
    • one year ago
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    but that's not one of my choices.

  5. Jhannybean
    • one year ago
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    What are your answer choices?

  6. anonymous
    • one year ago
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    ok \[f(g(x))= 2(x ^{2}-1)+1\]

  7. anonymous
    • one year ago
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    \[f(g(x))=2x ^{2}-2+1\]

  8. anonymous
    • one year ago
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    2(x^2-1)+1 = 2x^2 -2 + 1 = 2x^2 - 1

  9. Jhannybean
    • one year ago
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    Oh, I made a typo there.

  10. Jhannybean
    • one year ago
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    The \(x^2-2\) should have been \(x^2-1\)

  11. anonymous
    • one year ago
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    \[f(g(x))= 2x ^{2}-1\]

  12. anonymous
    • one year ago
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    hope this helps :)

  13. Jhannybean
    • one year ago
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    No harm in correcting me :)

  14. anonymous
    • one year ago
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    f(g(x)) = 2(g(x)) +1 = 2(x^2-1)+1

  15. sloppycanada
    • one year ago
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    So this one - Find g(f(x)) if f(x) = 2x - 3 and g(x) = x - 1.

  16. sloppycanada
    • one year ago
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    Should be - \[2x^2 + x -3\]

  17. anonymous
    • one year ago
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    g)x)-1 = (2x-3)-1 = 2x-4

  18. sloppycanada
    • one year ago
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    Find f(g(x)) if f(x) = 2x + 1 and g(x) = x2 - 1. My answer options are - 2x -1 \[2x^2-1\] \[2x^2 + 1\] \[2x^3-1\]

  19. sloppycanada
    • one year ago
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    So if it's f(g(x)), shouldn't it be something along the lines of - (x^2 - 1)(2x+1)?

  20. Jhannybean
    • one year ago
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    We've done this one.

  21. Jhannybean
    • one year ago
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    Just scroll up?

  22. sloppycanada
    • one year ago
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    It's just 2x -1 ?

  23. Jhannybean
    • one year ago
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    You're not multiplying anything actually, you're just substituting the function of g(x).

  24. sloppycanada
    • one year ago
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    Wait what? I am very confused.

  25. Jhannybean
    • one year ago
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    \[f(g(x))= f(x^2-1) = 2(x^2-1)+1 = 2x^2-2+1 =2x^2-1\]

  26. Jhannybean
    • one year ago
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    Do you see it now, @sloppycanada ?

  27. sloppycanada
    • one year ago
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    Kind of..

  28. Jhannybean
    • one year ago
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    which part is confusing you?

  29. sloppycanada
    • one year ago
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    Where the "x" in 2x-1 went.

  30. Jhannybean
    • one year ago
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    It was replaced by g(x)

  31. sloppycanada
    • one year ago
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    Okay, so in this - Find g(f(x)) if f(x) = 2x - 3 and g(x) = x - 1. It'd just be 1(2x -3) - 1?

  32. Jhannybean
    • one year ago
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    Not quite, \[g(f(x)) = g(2x-3)= (2x-3)-1 = 2x-4\]

  33. sloppycanada
    • one year ago
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    Okay so again - Find (f – g)(x) if f(x) = 2x – 6 and g(x) = 3x + 8. (2x -6) - (3x+ 8) = -x +2?

  34. Jhannybean
    • one year ago
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    Somewhat, but a small math error. \[(f-g)(x) = f(x) - g(x) = (2x-6)-(3x+8) = -x-(8+6) = -x-14\]

  35. Jhannybean
    • one year ago
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    You've got the idea though :D

  36. sloppycanada
    • one year ago
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    Find (f + g)(x) if f(x) = 1/2x + 4 and g(x) = -x - 3. (1/2x +4) + (-x-3) = -1/2 + 1

  37. sloppycanada
    • one year ago
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    -1/2x* +1?

  38. Jhannybean
    • one year ago
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    Yep

  39. sloppycanada
    • one year ago
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    Find the zeros of -4x + 6

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